This invention relates to an apparatus for characterizing physical properties of a test piece, and related methods.
In many circumstances, it is desirable to determine the physical properties of an object or structure. For example, in a manufacturing environment, it may be advantageous to determine selected physical properties of a product so as to verify that the product meets the minimum standards for its intended purpose. Similarly, it may be advantageous to determine the physical properties of a product in order to adjust a manufacturing process, so that a consistent and high-quality product may be produced. Furthermore, by evaluating the physical properties of a particular type of product while adjusting various process parameters, it is possible to optimize the process parameters for a particular product or process. For example, for the manufacture of a drawn polymer material, such parameters as the temperature of the molten material, the amount of draw-down, the through-put, and the viscosity of the base material may be determined and optimized through proper evaluation of physical properties of the finished product.
A wide variety of physical properties may be measured. One useful property is the elastic modulus. The elastic modulus is a ratio of the stress applied to an object or structure over strain exhibited by that object or structure.
Stress is defined as the applied force divided by the resisting area of the object. It will be appreciated that an object may be subjected to many sorts of stress, such as shear, compression, torsion, and tensile. It will also be appreciated that the resisting area of an object or structure depends on both the shape of the object and the type or direction of stress applied to it.
Strain is a relative deflection of the object in question in response to the applied stress. An object may similarly exhibit various sorts of strains depending on the stresses applied.
Regardless of the type of stress and strain, however, the elastic modulus may be understood to be an indication of an object""s response to applied forces.
One particularly useful elastic modulus is the Young""s modulus. The Young""s modulus is a ratio of the applied tensile stress xcex4 over the exhibited tensile strain xcex5. Although a wide variety of other elastic moduli may of course be measured, it is enlightening to consider as a specific example the Young""s modulus and devices known for measuring it.
Several devices are known for measuring Young""s modulus. Conventional devices include destructive (xe2x80x9ctest until failurexe2x80x9d) systems and dynamic mechanical analyzers, commonly known as DMAs. Neither type of device is entirely satisfactory, however.
Conventionally, the Young""s modulus of an object is determined destructively by measuring the deflection of the object while applying a gradually increasing force to the object until it xe2x80x9cfailsxe2x80x9d, that is, deforms plastically or fractures. The stress-strain relationship is then determined for the sample. The stress-strain relationship is plotted as a curve, and a straight-line fit of the curve is approximated. The slope of the line may be used as an approximation of Young""s modulus.
This approach has numerous disadvantages. For example, a useful approximation of Young""s modulus can only be determined if the stress-strain relationship is linear or nearly linear over a broad range. That is, the material must have a large and generally uniform range of elastic deformation. Although some materials have such properties, many others do not. In particular, many composite materials do not exhibit linear stress-strain relationships.
Also, the destruction of the object being tested is inherent in the test method. In order to obtain sufficient useful data for the straight-line fit, the sample must be tested across essentially its entire range of elastic deformation. Therefore, it is necessary to increase the applied force until the sample either deforms plastically or fractures. In either event, the sample is destroyed, and is unavailable for sale, further testing, or other purposes.
Because conventional destructive methods require a substantial range of stress-strain data to determine a useful value of Young""s modulus, destructive testing is generally not suitable for determining Young""s modulus at a particular stress, or for a narrow band of stresses. Conventional destructive testing typically produces only an average value for the entire range of applied stress.
In addition, destructive tests are generally limited to samples of standard size and shape. This is the case for several reasons. First, in order to calculate the stress on a sample, the resisting area of the sample must be known. Thus, in order to determine the Young""s modulus of an arbitrarily-shaped product using conventional destructive methods, the geometry of the product must be carefully measured, and the area calculated. For complex shapes, this is a considerable difficulty.
Second, the Young""s modulus of an object depends not only on its material and its resisting area but also on its shape. For example, a U-shaped beam generally exhibits a higher Young""s modulus than a flat strip, even if the strip and the beam have the same resisting area and are made from the same material. In many cases, especially for complex shapes, it is impractical or impossible to calculate in advance the effect of a particular shape on the Young""s modulus. Thus, in order to compare test subjects without performing complex corrections due to varying geometry, it is generally necessary to test a sample of standard size and shape instead of an actual product.
Although this simplification is convenient, the use of samples as opposed to actual products produces difficulties. For example, tests on samples tend to be inaccurate for orthotropic materials, that is, materials that have a directionally non-uniform structure. One common example is wood, which has a grain that is stronger in some directions than in others. Other orthotropic or partially orthotropic materials include composites, laminates, etc. Orthotropic materials pose difficulties for conventional tests for several reasons.
First, the stress and strain of orthotropic materials do not always vary linearly with increasing dimension. That is, doubling the cross-sectional area of a sample of an orthotropic material may not double the stress required to achieve a given strain, even if the material composition and shape are kept exactly the same.
Second, orientation is important for orthotropic materials. For example, an object composed of many laminated layers will have a very different Young""s modulus if the layers are oriented parallel to the direction of stress than if they are perpendicular to the direction of stress. Because it may be inconvenient or impossible to produce a test sample that is representative of the orientation in which the actual product will be used, the accuracy of the test becomes questionable at best.
Dynamic Mechanical Analyzers operate according to a different principle. As previously described, the Young""s modulus is a measure of material stiffness. The Young""s Modulus of an object, along with the object""s geometry, determine the stiffness coefficient of that object. In turn, the stiffness coefficient of an object determines the frequency at which it will vibrate. By proper analysis, it is therefore possible to determine the Young""s modulus of an object if the geometry and a frequency of vibration of the object are known.
In order to obtain data, the object must of course be made to vibrate. However, the vibration must be at the free vibrational frequency for the object, also known as its natural frequency. The free vibrational frequency of an object is the frequency at which it will vibrate if it is initially disturbed but not subsequently subjected to additional forces. Such vibrations are known as free vibrations.
If an object is driven at its free vibrational frequency, also known as its harmonic frequency, the object""s natural vibrations and the driving vibrations will combine additively, and the object is said to be vibrating harmonically. Harmonic vibration is generally detectable as an apparent increase in the amplitude of the object""s vibrations.
DMAs exploit this natural phenomenon. Driving vibrations are applied to an object to be tested. The frequency of the vibrations is slowly adjusted until the object undergoes harmonic vibration at its free vibrational frequency. At this point the vibrations are maintained at a constant, steady-state frequency, and the driving frequency, which at this point is known to be equal to the free vibrational frequency, of the object is measured. Young""s modulus is then calculated from the free vibrational frequency of the object.
However, DMAs also suffer from serious limitations.
First, as described above, the free vibrational frequency of an object depends not only on its Young""s modulus but also on its geometry. As described previously with respect to destructive testing, determining the geometry of an arbitrary object may be inconvenient or impossible. Therefore, for similar reasons, DMA testing is conventionally performed on samples of standard size and shape.
However, because of this, the problems inherent in testing standard samples as opposed to actual objects, as outlined above with respect to destructive testing, also apply to DMA testing.
In addition, DMA testing is very time-consuming. In order to obtain data, the test sample must undergo steady-state harmonic vibration. Thus, DMA testing requires that the driving frequency be changed at a rate that permits the harmonic vibration to reach a steady state and be recognized. In practice, this means that the frequency of the driving vibrations must be changed very slowly. Because the range of possible natural frequencies for a test piece is quite broad, considerable time is required to perform even a single DMA test.
Also, because of the need to precisely match an initially unknown frequency for every test sample, DMA testing requires a relatively high degree of skill and training to perform reliably. The need for a highly skilled operator limits the utility of DMA testing for controlling a manufacturing line from a production floor.
Furthermore, DMA testing is severely limited in terms of the amount of data generated. The stress-strain curve of any object is never perfectly linear. That is, the Young""s modulus of any object is different for different strains. For a vibration, the strain corresponds to the amplitude of the vibration. This means that the free vibrational frequency of an object varies at least slightly with the amplitude of the vibration. Because the harmonic vibrations necessary for DMA testing are steady-state, they permit measurement of only a single point of data on the stress-strain curve, and hence only a single value for Young""s modulus, per test. Varying the amplitude of the vibrations to determine a value for Young""s modulus at a different strain requires readjustment of the driving frequency to match the free vibrational frequency of the object for that different strain. Therefore, in order to generate a useful stress-strain curve, rather than a single approximating value, the test must be repeated many times. This can be inconvenient, especially in view of the relatively long time necessary for each individual test. In particular, the long time required renders DMA testing unsuitable for many process control applications, as it may not be possible to produce data quickly enough to permit timely adjustment of the process.
In contrast, the claimed invention is based on the observation of free vibrations in a system. Measurements of the free vibrations of the system may be used to conveniently determine physical properties of an object of an object of largely arbitrary geometry, without destroying the object being tested.
In order to appreciate the structure and function of the claimed invention, it is illustrative to consider certain properties of free vibration, and certain exemplary systems.
An object or system that is struck, deflected or otherwise disturbed and then left undisturbed undergoes free vibration. Free vibrations in an object may be represented as a single degree of freedom mass-spring-damper system 10, as illustrated in FIG. 1.
An ideal mass-spring-damper system 10 includes a mass 12, a spring 14 connecting the mass 12 to an effectively immobile base 16, and a damping element 18 also connecting the mass 12 to the base 16. In an equilibrium state, the spring 14 is neither compressed nor extended, and thus applies no net force to the mass 12. If the mass 12 is displaced by some initial distance, the spring 14 is either extended or compressed, and in either case a net force is a then applied to the mass 12. The mass 12 will then oscillate back and forth as the spring 14 alternately expands and contracts. At this point, the system 10 is undergoing free vibration.
For purposes of mathematical analysis, the physical properties of the system are defined as follows. The mass 12 has a possesses a mass M. The spring 14 has a stiffness coefficient K, also known as a spring coefficient. The damping element 16 has a damping coefficient C.
It is noted that the system illustrated in FIG. 1 is a one-dimensional system, wherein both the spring 14 and the damping element 16 are effectively massless, and wherein the mass 12 is a zero-dimensional point mass. For such a system, the entire mass M and only mass M have any effect on the system. Thus, the effective mass of the system is exactly equal to the mass of the mass 12. However, it will be appreciated by those knowledgeable in the art that for a real system in two or more dimensions, such as for example a longitudinal beam undergoing transverse longitudinal free vibrations, the mass M of a system as it relates to the following discussion would be an effective mass Meff, rather than simply an ideal point mass. For the sake of accuracy in the following discussion, the mass of the system will be referred to as the effective mass Meff. This matter is analyzed in greater detail below.
The stiffness coefficient K relates the force necessary to displace the mass M by a given distance, according to the equation:
F=Kyxe2x80x83xe2x80x83(Equation 1)
wherein
F is the force applied to the system;
K is the stiffness coefficient; and
y is the displacement of the mass.
Alternatively, Equation 1 may be rearranged to solve for K:                     K        =                  F          y                                    (                  Equation          ⁢                      xe2x80x83                    ⁢          2                )            
The displacement is also referred to as the deflection of the system, in particular for cases wherein the spring force is applied to bend or deflect a solid object such as a beam. It will be appreciated by those knowledgeable in the art that when an initial displacement is used to initiate free vibrations, the amplitude of the initial displacement is the same as the initial amplitude of the free vibrations.
The damping coefficient C is a measure of the forces opposing free vibrations in the system. In the simple case wherein the system is completely undamped, that is, the damping coefficient C is 0, the system vibrates according to the function:                               ω          n                =                              K                          M              eff                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          3                )            
wherein
xcfx89n is the natural frequency of vibrations of the system;
K is the stiffness coefficient of the spring; and
Meff is the effective mass.
For an undamped system the free vibration would continue indefinitely, at an amplitude exactly equal to the original displacement, and at a frequency determined by the effective mass and the stiffness coefficient according to Equation 3.
However, in any real system, energy is lost due to various damping factors, such as friction, etc. Regardless of the precise source the damping, energy is gradually lost from any freely vibrating system. Thus, the vibrations gradually decay until the system reaches static equilibrium and the free vibrations cease altogether.
The motion of a damped system is considerably more complex than that of an undamped system. For a spring-mass-damper system as illustrated in FIG. 1, the governing equation of motion may be written:                                                         M              eff                        ⁢                                                            ⅆ                  2                                ⁢                y                                            ⅆ                                  t                  2                                                              +                      C            ⁢                                          ⅆ                y                                            ⅆ                t                                              +          Ky                =                  f          ⁡                      (            t            )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          4                )            
wherein
Meff is the effective mass;
C is the damping coefficient;
K is the stiffness coefficient; and
y is the displacement of the mass.
It will be appreciated by those knowledgeable in the art that both the form of Equation 4 and the spring-mass-damper system illustrated in FIG. 1 describe linear systems, and that such systems are exemplary only. A wide variety of other damped free vibrational systems are possible, including but not limited to rotational spring-mass-damper systems. It will be further appreciated that the form of Equation 4 and the spring-mass-damper system illustrated in FIG. 1 represent one-dimensional systems, and that such systems are likewise exemplary only. Non-linear systems and systems undergoing free vibrations in two or more dimensions may be represented by equations similar to Equation 4. The mathematical analysis disclosed herein is similarly applicable to such alternative systems, and the discussion herein disclosed with respect to the exemplary system of FIG. 1 likewise applies equally to them.
Returning to the exemplary case of FIG. 1 and Equation 4, it will be appreciated that the rate at which the vibrations damp towards an amplitude of zero depends on the value of the damping coefficient C, that is, on the amount of damping present in the system. For any system, there is a quantity of damping for which the vibrations decay towards static equilibrium at the most rapid rate possible. This value of the damping coefficient is referred to as critical damping Cc. If the actual damping value is substantially lower than Cc the actual rate at which the vibrations approach an amplitude of zero is lower than it would be if the damping were Cc. If the actual damping value is substantially higher than Cc the system does not truly vibrate at all, but rather moves more or less steadily toward equilibrium.
A physical system may be characterized in terms of how closely the actual damping matches the critical damping value for that system. The damping ratio xcex6 is thus defined as                     ζ        ≡                  C                      C            c                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          5                )            
It will be appreciated that for a critically damped system, the damping ratio xcex6 is equal to 1.
It will further be appreciated that the value Cc for any spring-mass-damper system must depend on the effective mass Meff and the stiffness coefficient K. In mathematical terms, this dependence follows the equation:                               C          c                =                  2          ⁢                                                    M                eff                            ⁢              K                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          6                )            
Combining Equations 5 and 6 yields the following relation:                     ζ        =                  C                      2            ⁢                                                            M                  eff                                ⁢                K                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          7                )            
Equation 7 may also be solved for the damping coefficient C as:                     C        =                  2          ⁢          ζ          ⁢                                                    M                eff                            ⁢              K                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          8                )            
Returning now to the governing equation of motion as formulated in Equation 4, dividing all terms by K yields:                                                                         M                eff                            K                        ⁢                                                            ⅆ                  2                                ⁢                y                                            ⅆ                                  t                  2                                                              +                                    C              K                        ⁢                                          ⅆ                y                                            ⅆ                t                                              +          y                =                  f          ⁡                      (            t            )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          9                )            
As may be seen, by manipulation of terms and substitution of Equations 3 and 7 the following relation may be determined:                               C          K                =                                            (                              2                2                            )                        ⁢                          (                              C                                  (                                                            K                                        ⁢                                          K                                                        )                                            )                        ⁢                          (                                                                    M                    eff                                                                                        M                    eff                                                              )                                =                                    2              ⁢                              (                                                                            M                      eff                                                                            K                                                  )                            ⁢                              (                                  C                                      2                    ⁢                                                                                            M                          eff                                                ⁢                        K                                                                                            )                                      =                                          2                ⁢                ζ                                            ω                n                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          10                )            
In turn, substituting Equations 3 and 8 into Equation 9 yields:                                                         1                              ω                n                2                                      ⁢                                                            ⅆ                  2                                ⁢                y                                            ⅆ                                  t                  2                                                              +                                                    2                ⁢                ζ                                            ω                n                                      ⁢                                          ⅆ                y                                            ⅆ                t                                              +          y                =                  f          ⁡                      (            t            )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          11                )            
wherein
xcfx89n is the natural frequency of the system
y is the displacement of the system;
t is the time elapsed since the initiation of free vibrations; and
xcex6 is the damping ratio of the system.
Equation 11 may be usefully solved for y in the form:                     y        =                              ⅇ                          (                                                -                  ζ                                ⁢                                  xe2x80x83                                ⁢                                  ω                  n                                ⁢                t                            )                                ⁡                      (                                                            c                  1                                ⁢                                  sin                  ⁡                                      (                                                                  ω                        n                                            ⁢                                                                        1                          -                                                      ζ                            2                                                                                              ⁢                      t                                        )                                                              +                                                c                  2                                ⁢                cos                ⁢                                  xe2x80x83                                ⁢                                  (                                                            ω                      n                                        ⁢                                                                  1                        -                                                  ζ                          2                                                                                      ⁢                    t                                    )                                                      )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          12                )            
wherein
c1 and c2 are constants reflecting initial system conditions.
It is noted that c1 and c2 are not components of or otherwise directly related to the damping coefficient C.
It will be appreciated by those knowledgeable in the art that for physical reasons the frequency of free vibrations of a mass-spring-damper system where the damping coefficient (and hence the damping ratio) is non-zero is different from the frequency that would be observed for the same system if the damping coefficient were zero. The continuing loss of energy due to damping reduces the frequency of free vibration.
The actual frequency of free vibration for a spring-mass-damper system may be determined by inspection of Equation 12. For such an equation, the frequency is the expression associated with xe2x80x9ctxe2x80x9d upon which the trigonometric functions operate. Considering the case wherein the damping coefficient C and therefore the damping ratio xcex6 is zero, the expression       1    -          ζ      2      
equals exactly 1. Hence, Equation 12 for that special case could be written
y=e(xe2x88x92xcex6xcfx89nt)(c1 sin(xcfx89nt)+c2 cos(xcfx89nt))xe2x80x83xe2x80x83(Equation 13)
In such a case, the frequency is xcfx89n, as was previously stated. Similarly,
in Equation 12, the frequency with a non-zero damping coefficient may be extracted as:                               ω          d                =                              ω            n                    ⁢                                    1              -                              ζ                2                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          14                )            
wherein xcfx89d is the damped natural frequency of the system.
Alternately, Equation 14 may be solved in terms of xcfx89n as                               ω          n                =                              ω            d                                              1              -                              ζ                2                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          15                )            
It will be appreciated that the period associated with the damped natural frequency xcfx89d of the system is:                               T          d                =                              2            ⁢            π                                ω            d                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          16                )            
wherein
Td is the damped natural period of the system.
Returning to Equation 12, by substitution of Equation 14 therein, it can be written as:
y=e(xe2x88x92xcex6xcfx89nt)(c1 sin(xcfx89dt)+c2 cos(xcfx89dt))xe2x80x83xe2x80x83(Equation 17)
Alternatively, the solution of Equation 12 may be written in a different but equivalent form:
y=c3e(xe2x88x92xcex6xcfx89nt)sin(xcfx89dt+xcfx86)xe2x80x83xe2x80x83(Equation 18)
wherein
c3 is a constant reflecting initial system conditions;
xcfx86 is a phase constant reflecting initial system conditions.
It is noted that c3, like c1 and c2, is not a component of or otherwise directly related to the damping coefficient C.
As previously noted, once free vibrations are initiated in a mass-spring-damper system, energy is gradually lost to the damper and the vibrations decay. The relative amplitude of the peaks of two vibrations may be calculated by a simple ratio of Equation 18 for two values of n. It will be appreciated that the ratio of peak values is exemplary only. Although it is convenient in certain applications to compare one peak to another, comparing waves at other phases is also mathematically possible. Such alternative wave comparisons may be equally suitable for certain applications, and may be handled similarly.
It is convenient to consider the case where the first value of n is zero, and the second remains an arbitrary value n. It will be appreciated that in such a case, t for the peak at n=0 will be 0, and t for the peak at n will be Tdn. Therefore:                                           y            0                                y                          0              +              n                                      =                                                            c                3                            ⁢                              ⅇ                                  (                                                            -                                              ζω                        n                                                              ⁢                    t                                    )                                            ⁢                              sin                ⁡                                  (                                                                                    ω                        d                                            ⁢                      0                                        +                    φ                                    )                                                                                    c                3                            ⁢                              ⅇ                                  (                                                                                    -                                                  ζω                          n                                                                    ⁢                      t                                        +                                                                  T                        d                                            ⁢                      n                                                        )                                            ⁢                              sin                ⁡                                  (                                                                                    ω                        d                                            ⁢                                              T                        d                                            ⁢                      n                                        +                    φ                                    )                                                              =                                                    ⅇ                                                      -                                          ζω                      n                                                        ⁢                  t                                                            ⅇ                                                                            -                                              ζω                        n                                                              ⁢                    t                                    +                                                            T                      d                                        ⁢                    n                                                                        =                          ⅇ                                                ζω                  n                                ⁢                                  T                  d                                ⁢                n                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          19                )            
Taking the natural log of both sides yields the expression:                               l          ⁡                      (                                          y                0                                            y                                  0                  +                  n                                                      )                          =                              l            ⁢                          xe2x80x83                        ⁢                          ⅇ                                                ζω                  n                                ⁢                                  T                  d                                ⁢                n                                              =                                    ζω              n                        ⁢                          T              d                        ⁢            n                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          20                )            
Equation 20 can be solved in terms of xcex6 as:                     ζ        =                              l            ⁡                          (                                                y                  0                                                  y                                      0                    +                    n                                                              )                                                          ω              n                        ⁢                          T              d                        ⁢            n                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          21                )            
With substitution from Equations 14 and 16, Equation 21 may in turn be expressed as:                     ζ        =                                            l              ⁡                              (                                                      y                    0                                                        y                                          0                      +                      n                                                                      )                                                                    ω                n                            ⁢                              T                d                            ⁢              n                                =                                                    l                ⁡                                  (                                                            y                      0                                                              y                                              0                        +                        n                                                                              )                                                                                                  ω                    n                                    ⁡                                      (                                                                  2                        ⁢                        π                                                                    ω                        d                                                              )                                                  ⁢                n                                      =                                                            l                  ⁡                                      (                                                                  y                        0                                                                    y                                                  0                          +                          n                                                                                      )                                                                                                              ω                      n                                        ⁡                                          (                                                                        2                          ⁢                          π                                                                                                      ω                            n                                                    ⁢                                                                                    1                              -                                                              ζ                                2                                                                                                                                                        )                                                        ⁢                  n                                            =                                                                    1                    -                                          ζ                      2                                                                      ⁢                                                      l                    ⁡                                          (                                                                        y                          0                                                                          y                                                      0                            +                            n                                                                                              )                                                                            2                    ⁢                    π                    ⁢                                          xe2x80x83                                        ⁢                    n                                                                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          22                )            
Equation 22 represents an exact and general solution for any value of the damping ratio xcfx89. However, in practice, many systems of interest have a damping ratio xcfx89 that is substantially less than 1: It will be appreciated by those knowledgeable in the art that if the damping ratio xcfx89 substantially less than 1, the value of the expression       1    -          ζ      2      
closely approximates 1.
Using this approximation, Equation 22 may be simplified to the form:                     ζ        =                              l            ⁡                          (                                                y                  0                                                  y                                      0                    +                    n                                                              )                                            2            ⁢            π            ⁢                          xe2x80x83                        ⁢            n                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          23                )            
Based on the above mathematical derivations, it is possible to modify and substitute various of the preceding equations into Equation 6 such that the damping coefficient C is expressed in terms of directly measurable variables. First, Equation 3 is substituted into Equation 8:                     C        =                              2            ⁢            ζ            ⁢                                                            M                  eff                                ⁢                K                                              =                                    2              ⁢              ζ              ⁢                              xe2x80x83                            ⁢              M              ⁢                                                K                                      M                    eff                                                                        =                          2              ⁢              ζ              ⁢                              xe2x80x83                            ⁢                              M                eff                            ⁢                              ω                n                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          24                )            
Then Equation 24 is rearranged into equivalent forms:                     C        =                              2            ⁢            ζ            ⁢                          xe2x80x83                        ⁢                          M              eff                        ⁢                          ω              n                                =                                                    2                ⁢                ζ                ⁢                                  xe2x80x83                                ⁢                                  M                  eff                                                            1                                  ω                  n                                                      =                                                            4                  ⁢                  πζ                  ⁢                                      xe2x80x83                                    ⁢                                      M                    eff                                                                                        2                    ⁢                    π                                                        ω                    n                                                              =                                                                    (                                                                  4                        ⁢                        π                        ⁢                                                  xe2x80x83                                                ⁢                                                  M                          eff                                                                                            (                                                                              2                            ⁢                            π                                                                                ω                            n                                                                          )                                                              )                                    ⁢                                      1                                                                                            1                          -                                                      ζ                            2                                                                                                                                                1                          -                                                      ζ                            2                                                                                                                                ⁢                  ζ                                =                                                      (                                                                  4                        ⁢                        π                        ⁢                                                  xe2x80x83                                                ⁢                                                  M                          eff                                                                                            (                                                                              2                            ⁢                            π                                                                                                              ω                              n                                                        ⁢                                                                                          1                                -                                                                  ζ                                  2                                                                                                                                                                    )                                                              )                                    ⁢                                      (                                          ζ                                                                                                    1                            -                                                                          ⁢                        ζ                                                                                                                                                    (                  Equation          ⁢                      xe2x80x83                    ⁢          25                )            
Next, Equation 14 is substituted into Equation 25:                     C        =                                            (                                                4                  ⁢                  π                  ⁢                                      xe2x80x83                                    ⁢                                      M                    eff                                                                    (                                                            2                      ⁢                      π                                                                                      ω                        n                                            ⁢                                                                        1                          -                                                      ζ                            2                                                                                                                                )                                            )                        ⁢                          (                              ζ                                                      1                    -                                          ζ                      2                                                                                  )                                =                                    (                                                4                  ⁢                  π                  ⁢                                      xe2x80x83                                    ⁢                                      M                    eff                                                                    (                                                            2                      ⁢                      π                                                              ω                      n                                                        )                                            )                        ⁢                          (                              ζ                                                      1                    -                                          ζ                      2                                                                                  )                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          26                )            
Then, Equation 16 is substituted into Equation 26:                     C        =                                            (                                                4                  ⁢                  π                  ⁢                                      xe2x80x83                                    ⁢                                      M                    eff                                                                    (                                                            2                      ⁢                      π                                                              ω                      n                                                        )                                            )                        ⁢                          (                              ζ                                                      1                    -                                          ζ                      2                                                                                  )                                =                                                    4                ⁢                π                ⁢                                  xe2x80x83                                ⁢                                  M                  eff                                                            T                d                                      ⁢                          (                              ζ                                                      1                    -                                          ζ                      2                                                                                  )                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          27                )            
Finally, Equation 23 is substituted into Equation 28:                     C        =                                                            4                ⁢                π                ⁢                                  xe2x80x83                                ⁢                                  M                  eff                                                            T                d                                      ⁢                          (                              ζ                                                      1                    -                                          ζ                      2                                                                                  )                                =                                    (                                                4                  ⁢                  π                  ⁢                                      xe2x80x83                                    ⁢                                      M                    eff                                                                    T                  d                                            )                        ⁢                          (                                                                    ln                    ⁡                                          (                                                                        y                          0                                                                          y                                                      0                            +                            n                                                                                              )                                                                            2                    ⁢                    π                    ⁢                                          xe2x80x83                                        ⁢                    n                                                                                        1                    -                                                                  (                                                                              ln                            ⁡                                                          (                                                                                                y                                  0                                                                                                  y                                                                      0                                    +                                    n                                                                                                                              )                                                                                                            2                            ⁢                            π                            ⁢                                                          xe2x80x83                                                        ⁢                            n                                                                          )                                            2                                                                                  )                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          29                )            
C is the damping coefficient;
Meff is the effective mass;
Td is the damped natural period of the system;
Y0 is the peak displacement of the system after zero free vibrations;
n is the number of free vibrations; and
y0+n is the peak displacement of the system after n free vibrations.
It is similarly possible to calculate a value of the stiffness coefficient K for a system undergoing free vibrations without depending on the geometry of the system. Returning to Equation 3, and solving it in terms of K yields the relation:
K=xcfx89n2Mxe2x80x83xe2x80x83(Equation 30)
wherein
K is the stiffness coefficient of the spring
xcfx89n2 is the natural frequency of vibrations; and
Meff is the effective mass.
Substitution of the expression for xcfx89n from Equation 15 results in the relation:                     K        =                                                            (                                                      ω                    d                                                                              1                      -                                              ζ                        2                                                                                            )                            2                        ⁢                          xe2x80x83                        ⁢                          M              eff                                =                                                    ω                d                2                            ⁢                              M                eff                                                    1              -                              ζ                2                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          31                )            
Equation 16 may be rearranged to solve for the damped natural frequency xcfx89d:                               ω          d                =                              2            ⁢            π                                T            d                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          32                )            
Substituting Equation 32 into Equation 31 results in the relation:                     K        =                                                                              (                                                            2                      ⁢                      π                                                              T                      d                                                        )                                2                            ⁢                              M                eff                                                    1              -                              ζ                2                                              =                                    4              ⁢                              π                2                            ⁢                              M                eff                                                                    T                d                2                            ⁡                              (                                  1                  -                                      ζ                    2                                                  )                                                                        (                  Equation          ⁢                      xe2x80x83                    ⁢          33                )            
Substitution of the expression for xcex6 from Equation 22 yields:                     K        =                              4            ⁢                          π              2                        ⁢                          M              eff                                                                          T                d                2                            (                              1                -                                                      l                    ⁡                                          (                                                                        y                          0                                                                          y                                                      0                            +                            n                                                                                              )                                                                            2                    ⁢                    π                    ⁢                                          xe2x80x83                                        ⁢                    n                                                              )                        2                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          34                )            
wherein
K is the stiffness coefficient;
Meff is the effective mass;
Td is the damped natural period of the system;
Y0 is the peak displacement of the system after zero free vibrations;
Y0+n is the peak displacement of the system after n free vibrations; and
n is the number of free vibrations.
As was pointed out with respect to Equation 1, the force necessary to displace any spring-mass-damper system is a function of the stiffness coefficient and the displacement. In addition, in many useful cases, the force necessary to displace or deflect a system of a particular type is known or may be calculated based on the structure of that system.
It is useful to consider the exemplary case of a spring-mass-damper system having a beam undergoing a transverse longitudinal deflection. Two exemplary systems of this type may be seen in FIGS. 2-5.
It will be appreciated by those knowledgeable in the art that for a beam system as described above, the force necessary to cause a deflection at a particular point along the length of the beam depends in part on the location of the point or points at which the beam is secured, and likewise on the location of the point for which the deflection of the beam is to be established. FIG. 2 illustrates an exemplary system 20 having a longitudinal beam 22, fixed at a first end 24. The beam 22 has a center of mass 26. FIG. 3 illustrates the beam 22 with a second end 28 displaced. The force that must be applied to the second end 28 to establish a deflection of the beam 22 at the second end 28 may be determined according to the relation:                     F        =                                            3              ⁢              EI                                      L              3                                ⁢          y                                    (                  Equation          ⁢                      xe2x80x83                    ⁢          35                )            
wherein
F is the applied force;
E is the elastic modulus or Young""s modulus of the beam
I is moment of inertia of the beam;
L is the distance between the fixed point and the point at which deflection is determined; and
y is the distance of the beam""s deflection at the second end.
It will be appreciated by those knowledgeable in the art that for a beam system as illustrated in FIGS. 2 and 3, L is equal to the length of the beam. That is, the first end 24 is fixed, and the second end 28 is deflected, so that the L is the distance between them, the full length of the beam. Such a configuration is convenient for certain applications. However, it will be appreciated that this configuration is exemplary only, and that the displacement may be measured at essentially any point along the length of the beam. Consequently, L is not necessarily equal to the length of the beam for all suitable systems.
FIG. 4 illustrates a system 30 including a longitudinal beam 32, fixed at a first end 34 and at a second end 38. The beam has a center of mass 36. FIG. 5 illustrates the beam 32 with the center of mass 36 of the beam 32 displaced. The force that must be applied to the center of mass 36 to establish a deflection at the center of mass 36 may be determined according to the relation:                     F        =                                            192              ⁢              EI                                      L              3                                ⁢          y                                    (                  Equation          ⁢                      xe2x80x83                    ⁢          36                )            
wherein
y is the distance of the beam""s deflection at the center of mass.
As noted previously, L is not necessarily equal to the length of the beam for all suitable systems. For a system wherein there are two or more fixed points, L is determined from the distance to the nearest fixed point. It will be appreciated by those knowledgeable in the art that for a beam system as illustrated in FIGS. 4 and 5, L is equal to half the length of the beam.
The values 3 and 192 in Equations 35 and 36 are geometric coefficients that correspond to the arrangement of the beam within the system. It will be appreciated by those knowledgeable in the art that the choice of a point on the beam at which to establish the displacement of the beam may be essentially arbitrary. The end points and the center of mass are often used as a matter of convenience, but other locations on the beam may be equally suitable. For other locations or for other systems based on a longitudinal beam, coefficients other than those in Equations 35 and 36 are applicable. In general, for a longitudinal beam, the equation of force will be of the form:                     F        =                                                            k                xe2x80x3                            ⁢              EI                                      L              3                                ⁢          y                                    (                  Equation          ⁢                      xe2x80x83                    ⁢          37                )            
wherein
kxe2x80x3 is a geometric coefficient corresponding to the arrangement of the beam. It is noted that although kxe2x80x3 is dependent on the arrangement of the beam as a whole within the test system, kxe2x80x3 is not dependent on the internal geometry of the beam.
In physical terms, the value of kxe2x80x3 depends on the amount of deflection that is enabled for a given system at a given force. It will be appreciated by those knowledgeable in the art that a beam fixed at both ends requires substantially more force to deflect than a beam that is fixed at one end. The differing values of kxe2x80x3 reflect this physical difference.
It will be appreciated by those knowledgeable in the art that the choice of a point on the beam at which to establish the displacement of the beam may be essentially arbitrary. The end points and the center of mass are often used as a matter of convenience, but other locations on the beam may be equally suitable.
It is also noted that kxe2x80x3 is not a component of or otherwise directly related to the stiffness coefficient K.
Methods for calculating the applicable coefficient kxe2x80x3 are well known, and are not described further herein. Similarly, methods for calculating the necessary force in general for a given displacement in systems other than beams is also well-known, and are not described further herein.
The moment of inertia I of any structure is dependent on the distribution of mass and on the geometry of the point about which the object is to be moved. It will be appreciated by those knowledgeable in the art that the internal structure is relevant to the moment of inertia only in so far as it affects the mass distribution. Thus, so long as the density is constant with respect to the motion, the internal structure of the system has no direct effect on the moment of inertia.
For example, for a beam as in FIGS. 2-5, so long as the distribution of mass of the beam is constant along the beam""s length, it is not necessary to know the cross-sectional structure of the beam. A solid cylindrical beam and a hollow square beam of equal length and equal mass per unit length have exactly the same moment of inertia about their end points. Thus, for beams of constant linear density, the internal structure need not even be determined in order to calculate the moment of inertia.
Combination of Equations 1 and 37 yields the following relation:                     Ky        =                                                            k                xe2x80x3                            ⁢              EI                                      L              3                                ⁢          y                                    (                  Equation          ⁢                      xe2x80x83                    ⁢          38                )            
Dividing both sides of Equation 38 by y and solving for E results in:                     E        =                              KL            3                                              k              xe2x80x3                        ⁢            I                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          39                )            
Substituting the expression for K from Equation 34 into Equation 39 gives:                     E        =                                            (                                                4                  ⁢                                      π                    2                                    ⁢                                      M                    eff                                                                                                              T                      d                      2                                        (                                          1                      -                                                                        l                          ⁡                                                      (                                                                                          y                                0                                                                                            y                                                                  0                                  +                                  n                                                                                                                      )                                                                                                    2                          ⁢                          π                          ⁢                                                      xe2x80x83                                                    ⁢                          n                                                                                      )                                    2                                            )                        ⁢                          L              3                                                          k              xe2x80x3                        ⁢            I                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          40                )            
Equation 40 may be simplified into the form:                     E        =                              4            ⁢                          π              2                        ⁢                          M              eff                        ⁢                          L              3                                                          k              xe2x80x3                        ⁢                          T              d              2                        ⁢                                          I                (                                  1                  -                                                            l                      ⁡                                              (                                                                              y                            0                                                                                y                                                          0                              +                              n                                                                                                      )                                                                                    2                      ⁢                      π                      ⁢                                              xe2x80x83                                            ⁢                      n                                                                      )                            2                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          41                )            
wherein
E is the Young""s modulus of the beam;
L is the distance between the fixed point and the point at which deflection is determined;
kxe2x80x3 is a geometric coefficient corresponding to the arrangement of the beam;
Meff is the effective mass;
Td is the damped natural period of the system;
I is moment of inertia of the beam;
y0 is the peak displacement of the system after zero free vibrations;
Y0+n is the peak displacement of the system after n free vibrations; and
n is the number of free vibrations.
As previously noted, for a real system as opposed to a zero-dimensional ideal mass-damper-system, the effective mass Meff will not necessarily be equal to the simple total mass of the system. For a beam system fixed at one end and deflected at the other end as illustrated in FIGS. 2 and 3, the effective mass Meff may be described according to the relation:
Meff=0.2357Mtestxe2x80x83xe2x80x83(Equation 42)
wherein
Mtest is the measured mass of the beam.
Similarly, for a beam system fixed at both ends and deflected at the center of mass as illustrated in FIGS. 4 and 5, the effective mass Meff may be described according to the relation:
Meff=0.3610Mtestxe2x80x83xe2x80x83(Equation 43)
The values 0.2357 and 0.3610 in Equations 42 and 43 are geometric coefficients that correspond to the relative amount of motion of the beam along its length, which depends in turn on the relative location of the fixing point or points along the length of the beam. In general, for a longitudinal beam, the effective mass will be of the form:
Meff=kxe2x80x2Mtestxe2x80x83xe2x80x83(Equation 44)
wherein
kxe2x80x2 is a geometric coefficient corresponding to the location along the length of the beam of the point at which force is applied.
In physical terms, the value of kxe2x80x2 depends on the aggregate deflection of the beam when force is applied at a given point. It will be appreciated by those knowledgeable in the art that when a beam that is fixed at a point or points is deflected, the fixed points of the beam do not deflect at all. Thus, a deflection of amplitude y at any particular point on the beam does not imply that the entire beam has moved a distance y. This variation in the deflection of various parts of the beam may be accounted for by use of the coefficient kxe2x80x2. In effect, kxe2x80x2 changes the effective mass of the beam. Although physically the mass of the beam is constant while the relative displacement during free vibration varies with position along the length of the beam, it is mathematically convenient and functionally equivalent to treat the system as though the displacement of the beam is uniform along its length while effectively altering the mass of the beam.
The relative motion of the system as a whole depends in part on the number and location of the fixed points. This physical difference is reflected accounted for with differing values of kxe2x80x2.
It is noted that kxe2x80x2 is not a component of or otherwise directly related to the stiffness coefficient K.
Methods for calculating the value of the coefficient kxe2x80x2 are well known, and are not described further herein.
It will be appreciated by those knowledgeable in the art that 0.2357 and 0.3610, the exemplary values of kxe2x80x2 in Equations 42 and 43, are approximations. However, the value of kxe2x80x2 may be calculated to any arbitrary precision.
It will also be appreciated by those knowledgeable in the art that as a practical matter, when observing an actual spring-mass-damper system, it may be convenient to attach a sensor to the system in order to facilitate measurement of the displacement during free vibration. For example, in the case of the beam systems shown in FIGS. 2-5, it may be convenient to attach a sensor to the beam. It will be appreciated that the additional mass of a sensor will change the effective mass of the system. This may be accounted for by adding a term to Equation 44:
Meff=kMsensor+kxe2x80x2Mtestxe2x80x83xe2x80x83(Equation 45)
wherein
k is a geometric coefficient corresponding to the location of the sensor""s center of mass along the length of the beam; and
Msensor is the mass of the sensor.
It will be appreciated that a sensor placed at a fixed point of a system would not contribute any effective mass to the system, since it would not move. In such a case, the value of k would be 0. Contrariwise, a sensor placed at the point at which displacement is a maximum would contribute its entire mass to the system as effective mass. In such a case, the value of k would be 1.
Although the actual mass of a sensor does not depend on its location, the system may be treated and analyzed as though this were the case by the use of the coefficient k to adjust the contribution of the sensor to the system""s total effective mass.
It is noted that k is not a component of or otherwise directly related to the stiffness coefficient K.
It will be appreciated by those knowledgeable in the art that for certain systems, including but not limited to the beam systems illustrated in FIGS. 2-5, it is convenient to locate the sensor such that k has a value of 1. For a beam system fixed at one end as illustrated in FIGS. 2-3, the sensor would be located at the second end 28. For a beam system fixed at both ends as illustrated in FIGS. 4-5, the sensor would be located at the beam""s center of mass 36. However, although such arrangements may be convenient for certain applications, it will be appreciated that they are exemplary only, and that the sensor could be placed in other locations. The value of k may be calculated for any arbitrary location of the sensor""s center of mass.
Methods for calculating the value of the coefficient k are well known, and are not described further herein.
Substitution of Equation 45 into Equations 29, 34, and 41 yields the following relations:                     C        =                              (                                          4                ⁢                                  π                  ⁡                                      (                                                                  kM                        sensor                                            +                                                                        k                          xe2x80x2                                                ⁢                                                  M                          test                                                                                      )                                                                              T                d                                      )                    ⁢                      (                                                            ln                  ⁡                                      (                                                                  y                        0                                                                    y                                                  0                          +                          n                                                                                      )                                                                    2                  ⁢                  π                  ⁢                                      xe2x80x83                                    ⁢                  n                                                                              1                  -                                                            (                                                                        ln                          ⁡                                                      (                                                                                          y                                0                                                                                            y                                                                  0                                  +                                  n                                                                                                                      )                                                                                                    2                          ⁢                          π                          ⁢                                                      xe2x80x83                                                    ⁢                          n                                                                    )                                        2                                                                        )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          46                )            
wherein
C is the damping coefficient;
k is a geometric coefficient corresponding to sensor location;
Msensor is the mass of the sensor;
kxe2x80x2 is a geometric coefficient corresponding to the location along the length of the beam of the point at which force is applied;
Mtest is the mass of the beam;
Td is the damped natural period of the system;
y0 is the peak displacement of the system after zero free vibrations;
n is the number of free vibrations; and
Y0+n is the peak displacement of the system after n free vibrations.                     K        =                              4            ⁢                                          π                2                            ⁡                              (                                                      kM                    sensor                                    +                                                            k                      xe2x80x2                                        ⁢                                          M                      test                                                                      )                                                                        T              d              2                        (                          1              -                                                (                                                            ln                      ⁡                                              (                                                                              y                            0                                                                                y                                                          (                                                              0                                +                                n                                                            )                                                                                                      )                                                                                    2                      ⁢                      π                      ⁢                                              xe2x80x83                                            ⁢                      n                                                        )                                2                                      )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          47                )            
K is the stiffness coefficient;
k is a geometric coefficient corresponding to sensor location;
Msensor is the mass of the sensor;
kxe2x80x2 is a geometric coefficient corresponding to the location along the length of the beam of the point at which force is applied;
Mtest is the mass of the beam;
Td is the damped natural period of the system;
y0 is the peak displacement of the system after zero free vibrations;
y0+n is the peak displacement of the system after n free vibrations; and
n is the number of free vibrations.                     E        =                              4            ⁢                          π              2                        ⁢                                          L                3                            ⁡                              (                                                      kM                    sensor                                    +                                                            k                      xe2x80x2                                        ⁢                                          M                      test                                                                      )                                                                        k              xe2x80x3                        ⁢                          T              d              2                        ⁢                          I              (                              1                -                                                      (                                                                  ln                        ⁡                                                  (                                                                                    y                              0                                                                                      y                                                              (                                                                  0                                  +                                  n                                                                )                                                                                                              )                                                                                            2                        ⁢                        π                        ⁢                                                  xe2x80x83                                                ⁢                        n                                                              )                                    2                                            )                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          48                )            
E is the Young""s modulus of the beam;
L is the distance between the fixed point and the point at which deflection is determined;
k is a geometric coefficient corresponding to sensor location;
Msensor is the mass of the sensor;
kxe2x80x2 is a geometric coefficient corresponding to the location along the length of the beam of the point at which force is applied;
Mtest is the mass of the beam;
kxe2x80x3 is a geometric coefficient corresponding to the arrangement of the beam;
Td is the damped natural period of the system;
I is moment of inertia of the beam;
y0 is the peak displacement of the system after zero free vibrations;
y0+n is the peak displacement of the system after n free vibrations; and
n is the number of free vibrations.
It will be appreciated by those knowledgeable in the art that although the beam systems illustrated in FIGS. 2-5 are analyzed mathematically herein, they are exemplary only. A wide variety of other systems exhibiting free vibrations may be equally suitable.
For example, for certain applications it may be advantageous to initiate transverse longitudinal free vibrations in a beam fixed at its center of mass. Such a system is illustrated in FIGS. 10 and 11. FIG. 10 illustrates an exemplary system 50 having a longitudinal beam 52, fixed at its center of mass 56. The first and second ends 54 and 58 are free to move. FIG. 11 illustrates the beam 52 with the first and second ends 54 and 58 displaced.
Such a system may be analyzed mathematically in a fashion similar to that disclosed herein with respect to the systems illustrated in FIGS. 2-5. However, for the sake of brevity, the system of FIGS. 10 and 11 is not analyzed mathematically herein. Similarly, systems with test pieces other than longitudinal beams are not analyzed mathematically herein. However, it will be appreciated that in addition to those systems specifically analyzed herein, other systems utilizing longitudinal beams as well as systems utilizing test pieces with configurations other than longitudinal beams may be equally suitable.
It will be appreciated by those knowledgeable in the art that for a real system, as opposed to the one-dimensional ideal system illustrated in FIG. 1, free vibration is possible in multiple directions and orientations, or modes. With regard to multiple modes of free vibration, it is again illustrative to consider the exemplary beam system of FIGS. 2 and 3.
As previously described, FIGS. 2 and 3 illustrate an exemplary system 20 including a beam 22 fixed at a first end. FIG. 2 illustrates the beam 22 in an undeflected or neutral position. FIG. 3 illustrates the beam 22 with a transverse longitudinal deflection, as would be observed during transverse longitudinal free vibration.
FIGS. 6 and 7 illustrate the same system 20 from the perspective of the second end 28. FIG. 6 illustrates the beam 22 in a neutral position. FIG. 7 illustrates the beam 22 with a torsional deflection, as would be observed during torsional free vibration. As shown, the second end 28 of the beam 22 is twisting about an axis 40 running longitudinally through the beam 22.
It will be appreciated by those knowledgeable in the art that the vibratory modes illustrated in FIGS. 2, 3, 6 and 7 are exemplary only, and that additional vibratory modes beyond transverse longitudinal and torsional, including but not limited to compressional, are possible. In addition, although the vibratory modes illustrated in FIGS. 2, 3, 6, and 7 are shown individually for purposes of clarity, it will be appreciated that it is possible for a single system to vibrate simultaneously in multiple modes.
It will also be appreciated that, although the longitudinal beam illustrated in FIGS. 2, 3, 6, and 7 has a simple rectangular cross-section, this configuration is exemplary only. The principles of the claimed invention are equally applicable to beams of essentially arbitrary cross-section, and to objects other than longitudinal beams, as is elsewhere noted herein.
It will further be appreciated that a given system will not necessarily have the same free vibrational properties in all vibratory modes.
For example, it will be appreciated by those knowledgeable in the art that the effective mass of a system will not necessarily be the same for different vibratory modes. This is because the relative amount of motion of different parts of the system varies with the mode of vibration. For example, referring again to FIGS. 2, 3, 6, and 7 it will be appreciated that the effective mass of the beam 42 for the transverse longitudinal free vibrations illustrated in FIGS. 2 and 3 will not necessarily be the same as the effective mass of the beam 42 for the torsional free vibrations illustrated in FIGS. 6 and 7.
Similarly, the damping coefficient, stiffness coefficient, and Young""s modulus of a given system will not necessarily be equal for different modes of vibration.
For a real system, therefore, Equations 29, 34, and 41 provide values of C, K, and E that are particular to a single mode of vibration. For example, for a beam system as those illustrated in FIGS. 2-5, the mode of vibration under consideration is transverse longitudinal vibration. Thus, Equations 29, 34, and 41 may be more specifically written as                               C          L                =                              (                                          4                ⁢                                  π                  (                                                            kM                      sensor                                        +                                                                  k                        xe2x80x2                                            ⁢                                              M                        test                                                                                                                        T                d                                      )                    ⁢                      (                                                            ln                  ⁡                                      (                                                                  y                        0                                                                    y                                                  (                                                      0                            +                            n                                                    )                                                                                      )                                                                    2                  ⁢                  π                  ⁢                                      xe2x80x83                                    ⁢                  n                                                                              1                  -                                                            (                                                                        ln                          ⁡                                                      (                                                                                          y                                0                                                                                            y                                                                  (                                                                      0                                    +                                    n                                                                    )                                                                                                                      )                                                                                                    2                          ⁢                          π                          ⁢                                                      xe2x80x83                                                    ⁢                          n                                                                    )                                        2                                                                        )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          49                )            
wherein
CL is the transverse longitudinal damping coefficient.                               K          L                =                              4            ⁢                                          π                2                            ⁡                              (                                                      kM                    sensor                                    +                                                            k                      xe2x80x2                                        ⁢                                          M                      test                                                                      )                                                                        T              d              2                        (                          1              -                                                (                                                            ln                      ⁡                                              (                                                                              y                            0                                                                                y                                                          (                                                              0                                +                                n                                                            )                                                                                                      )                                                                                    2                      ⁢                      π                      ⁢                                              xe2x80x83                                            ⁢                      n                                                        )                                2                                      )                                              (                  Equation          ⁢                      xe2x80x83                    ⁢          50                )            
KL is the transverse longitudinal stiffness coefficient.                               E          L                =                              4            ⁢                          π              2                        ⁢                                          L                3                            ⁡                              (                                                      kM                    sensor                                    +                                                            k                      xe2x80x2                                        ⁢                                          M                      test                                                                      )                                                                        k              xe2x80x3                        ⁢                          T              d              2                        ⁢                          I              (                              1                -                                                      (                                                                  ln                        ⁡                                                  (                                                                                    y                              0                                                                                      y                                                              (                                                                  0                                  +                                  n                                                                )                                                                                                              )                                                                                            2                        ⁢                        π                        ⁢                                                  xe2x80x83                                                ⁢                        n                                                              )                                    2                                            )                                                          (                  Equation          ⁢                      xe2x80x83                    ⁢          51                )            
EL is the transverse longitudinal Young""s modulus of the beam;
It will be appreciated by those knowledgeable in the art that the values of k, kxe2x80x2, kxe2x80x3, Td, and I will also be particular to each mode of vibration of the system. For example, for the transverse longitudinal vibration of the systems illustrated in FIGS. 2-5 these factors could be identified as kL, kxe2x80x2L, kxe2x80x3L, TdL, and IL. However, to avoid unnecessarily complicating the notation, when solving for E, C, or K in a particular mode of vibration, the factors k, kxe2x80x2, kxe2x80x3, Td, and I are assumed herein to be the appropriate factors for that mode.
It is noted that Equation 49 permits calculation of the damping coefficient from a mass, a measured period, and two measured peak displacements. In particular, it is noted that Equation 49 does not depend on the shape or geometry of the test piece.
Likewise, it is noted that Equation 50 permits calculation of the stiffness coefficient from a measured mass, a measured period, and two measured peak displacements. In particular, it is noted that Equation 50 does not depend on the shape or geometry of the test piece.
It is noted that Equation 51 permits calculation of the stiffness coefficient from a measured mass, a measured period, a measured length, and two measured peak displacements. However, Equation 51 does not depend on the shape or geometry of the beam, so long as the density of the beam is constant along its length.
In view of the preceding, it is the purpose of the claimed invention to overcome the deficiencies of existing apparatuses and methods for determining non-musical physical properties of objects.
According to the principles of the claimed invention, free vibrations are initiated in a test piece and are observed as they decay in amplitude, without continued driving. Data from the free vibrations as they decay is then analyzed to determine one or more physical properties of the test piece. Testing according to the principles of the claimed invention therefore provides data regarding physical properties across a large dynamic range with only a single test. Testing is quick, simple, and non-destructive to the test piece. Furthermore, a sample of arbitrary geometry may be tested without a need for complex geometric analysis. Because of this, it is possible to test a piece of an actual product rather than a standard sample, and the inaccuracies inherent in testing samples are avoided.
It is also the purpose of the claimed invention to provide an apparatus for measuring at least one non-musical physical property of a test piece.
An embodiment of a test apparatus in accordance with the principles of the claimed invention includes a frame with a natural vibrational frequency that is different from the anticipated free vibrational frequency of the test piece. It also includes fixing means for fixing the test piece to the apparatus at at least one fixing point, so that the test piece may undergo free vibrations. The apparatus also includes initiating means to initiate vibrations within the test piece. The apparatus further includes displacement measuring means to measure the displacement of the test piece and time measuring means in communication with the displacement measuring means so as to measure the time-varying displacement of the test piece as it undergoes free vibration.
Another embodiment of a test apparatus in accordance with the principles of the claimed invention includes a mechanism for collecting measurements of displacement and time. The collecting mechanism may include a computer.
Another embodiment of a test apparatus in accordance with the principles of the claimed invention includes a mechanism for processing measurements of displacement and time so as to determine the physical properties of the test piece. The processing mechanism may include a computer.
Another embodiment of a test apparatus in accordance with the principles of the claimed invention includes a mechanism for recording measurements of displacement and time for later reference. The recording mechanism may include a computer.
Another embodiment of a test apparatus in accordance with the principles of the claimed invention includes a mechanism for displaying measurements of displacement and time. The measurements may be displayed in graphic form. The display mechanism may include a computer.
It is also the purpose of the claimed invention to provide a test method for measuring at least one non-musical physical property of a test piece.
An embodiment of a test method in accordance with the principles of the claimed invention includes the steps of fixing a test piece to a frame, initiating free vibrations in the test piece, measuring the time-varying displacement of the test piece as it undergoes free vibration, and calculating at least one value of the physical property of the test piece.
Another embodiment of a test method in accordance with the principles of the claimed invention further includes the step of displaying the values of the physical property. The values may be displayed on a computer.
Another embodiment of a test method in accordance with the principles of the claimed invention further includes the step of supplying feedback for adjusting a manufacturing process based on the values of the physical property.
It is also the purpose of the claimed invention to provide a method for controlling a manufacturing process based on measured values of a physical property of a test piece produced by the manufacturing process.
An embodiment of a process control method in accordance with the principles of the claimed invention includes the steps of fixing a test piece produced by the process to a frame, initiating free vibrations in the test piece, measuring the time-varying displacement of the test piece as it undergoes free vibration, calculating at least one value of the physical property of the test piece, and adjusting the process based on the value of the physical property.